a n b n ) n N is convergent with b n is convergent.

Transcription

1 32 Series Let s n be the n-th partial sum of n N and let t n be the n-th partial sum of n N. For k n we then have n n s n s k = a i a i = t n t k. i=k+1 i=k+1 Since t n n N is convergent by assumption, it is Cauchy. Now shows that also s n n N is Cauchy. So by 1.7.2, s n n N converges. Notice that if 0 for all n, then is absolutely convergent if and only if it is convergent. On the other hand, the alternating harmonic series is an example of a series which is convergent but not absolutely convergent cf The geometric series n=0 xn for x < 1 and the series 1 n are examples 2 of absolutely convergent series Observation. The statements of also holds true if we replace convergent by absolutely convergent everywhere. Exercise! Comparison test. If is absolutely convergent and b n n N is a sequence with b n for all n N, then b n is absolutely convergent. Proof. We have to show that b n is convergent. By it is enough to show that the partial sums of b n n N are bounded. However, as b n for all n, the n th partial sums of b n n N is less or equal to then th partial sums of n N. Now we may use again to conclude that the partial sums of n N, and therefore also the partial sums of b n n N, are bounded Corollary. If is an absolutely convergent series and b n n N is a bounded sequence, then the series b n is absolutely convergent. Proof. Take B R with b n B for all n. Then b n B for all n. Since is absolutely convergent, also B is absolutely convergent see Hence by 2.2.5, also b n is absolutely convergent. A special case of is called Limit comparison test. If,b n > 0 for all n N and bn n N is convergent with then b n lim 0, n is convergent b n is convergent. Proof. By symmetry and the division rule see 1.6.3, we only need to show the implication. So assume converges. As > 0 for all n, is absolutely convergent. Since bn n N is convergent, it is bounded. Hence by also bn is convergent, as desired. Here an application of the limit comparison test Example. The series n+4 2n 3 n+1 is convergent.

2 Convergence tests for series 33 Proof. We apply the limit comparison test to the given series and the convergent series 1 n 2 cf We have n+4 2n 3 n+1 1 n 2 = n3 +4n2 2n 3 n as n. Since 1 n is convergent by 2.1.6, the limit comparison test tells us that also 2 is convergent. n+4 2n 3 n Ratio test. If n N is a sequence with 0 for all n N such that the sequence an+1 n N is convergent with then is absolutely convergent. lim +1 < 1, n Proof. Let x < 1 such that lim n an+1 < x. For some N N, we then have an+1 < x whenever n N. It follows a N+1 < x a N a N+2 = a N+2 a N+1 a N+1 < x x a N = x 2 a N a N+3 = a N+3 a N+2 a N+2 < x x 2 a N = x 3 a N. a N+k+1 = a N+k+1 a N+k a N+k < x x k a N = x k+1 a N. Thus a N+k x k a N for all k N. Since x = x < 1, the geometric series k=1 xk is absolutely convergent. Hence by 2.2.5, also k=1 a N+k is absolute convergent and therefore the original series is absolutely convergent. Notice that the ratio test has a weak converse: Let n N be a sequence with 0 for all n such that the sequence an+1 n N is convergent with lim +1 > 1. n Then is divergent Exercise!. On the other hand, if lim n an+1 = 1 we cannot deduce whether is convergent or divergent. For example, if = 1 n, then lim n an+1 = 1 and 1 n is divergent; whereas if = 1 n, then lim 2 n an+1 = 1, too. But 1 n is convergent. So in this case 2 further investigations are required. There are many other and stronger tests for convergence. The interested reader is referred to Wilde s book.

3 34 Series Example. For every c R, the series c n n! n=0 is absolutely convergent. To see this we may assume that c 0 otherwise cn n! = 0 for n 0 and the series converges to 1. We apply the ratio test and find c n+1 n+1! c n n! c = n+1 0 as n The series in the example above is the exponential series and one can show that c n n! = ec, n=0 where e is the Euler number. In fact, depending whom you ask, the Euler number is defined as 1 e := n!. n=0

4 Definition and first examples Continuous functions 3.1. Definition and first examples. We shall study functions f : S R, where S is onempty subset of R. Most of the time S will be an interval, i.e. S will be of the form [a,b] = {r R a r b} a,b = {r R a < r < b} [a,b = {r R a r < b} a,b] = {r R a < r b} a,+ = {r R a < r} [a,+ = {r R a r},b = {r R r < b},b] = {r R r b},+ = R bounded closed interval bounded open interval half open/closed interval half open/closed interval unbounded open interval unbounded closed interval unbounded open interval unbounded closed interval An open interval is any interval of the form a,b, a,+,,a, or R itself Definition of continuity. Let S R and f : S R be a function. Pick a point x 0 S. Then f is continuous at the point x 0 if ε > 0 δ > 0 x S x x 0 < δ fx fx 0 < ε. The function f : S R is continuous if f is continuous at each point x 0 S Discussion of the definition of continuity and initial examples. Firstly we see that the property of f : S R being continuous is a local property, i.e. it has to be checked at every point x 0 S. a Here is another formulation of continuity at a point. The function f : S R is continuous at x 0 S if and only if the following condition holds true: For every open interval J containing fx 0, there is an open interval I containing x 0 with fi S J. To see this, suppose this condition holds true. We want to show that f is continuous at x 0. Let ε > 0. We must find some δ > 0 such that x S x x 0 < δ fx fx 0 < ε. We define the open interval J as J = fx 0 ε,fx 0 +ε. ThenJ containsfx 0 andbyassumption, thereisanopenintervali containing x 0 such that fs I J. Since I is open, there is some δ > 0 such that x 0 δ,x 0 +δ I. Since fs I J we then also have + fs x 0 δ,x 0 +δ J.

5 36 Continuous functions Now, condition + is exactly the same as condition convince yourself that these conditions say the same thing. Therefore is proved. Conversely, assume that f is continuous at x 0. Take an open interval J of R with fx 0 J. Since J is an open interval, there is some ε > 0 such that fx 0 ε,fx 0 + ε J. As f is continuous at x 0 there is some δ > 0 such that condition holds. We choose I = x 0 δ,x 0 + δ. Then I is an open interval, containing x 0 and condition precisely says that fs I fx 0 ε,fx 0 +ε. b The real number δ in depends on ε. This can be made transparent if we choose fx = x and S = R. Then continuity at x 0 S reads as ε > 0 δ > 0 x R x x 0 < δ x x 0 < ε. This condition clearly holds true: We simply choose δ = ε. On the way we have now also shown that the function fx = x is continuous no matter what S is. c Let us illustrate a and b with an example. Let f : R R be defined by fx = x 2 so here, S = R. We want to show that f is continuous at each x 0 R. Let ε > 0. We have to find δ > 0 depending on ε as required by definition Thus, δ has to be so that x 2 x 2 0 < ε whenever x x 0 < δ. This gives us a condition we can work with: We try to estimate x 2 x 2 0 in terms of x x 0 :We have x 2 x 2 0 = x x 0 x+x 0 = x x 0 x+x 0. Nomatterhowwechooseδ, weseethatif x x 0 < δ, then x 2 x 2 0 δ x+x 0. Hence if ε were so that δ x+x 0 < ε, then we are done. ε We choose 1 > δ > 0 such that δ < 1+ x. Then for x x 0+x 0 0 < δ we have x+x 0 = x x 0 +2x 0 x x 0 + 2x 0 < δ + 2x 0 and δ x+x 0 < ε 1+ x 0 +x 0 δ + x 0 +x 0 < ε as desired. This shows that fx = x 2 defined on R is continuous. A proof of continuity as above is called a proof from first principles. d A trivial but useful observation is the following: If f : S R is continuous and T S, then the restriction of f to T is continuous, too. Exercise! e Here an example of a function that is not continuous: The sign function sign : R R defined by 1 if x > 0 signx = 0 if x = 0 1 if x < 0 is not continuous.

6 Definition and first examples 37 We show that sign is not continuous at x 0 = 0. This means we have to show that ε > 0 δ > 0 x S x x 0 < δ and signx signx 0 ε We choose ε = 1 any choice smaller that 1 will also work!. This choice is motivated by the graph of the sign-function. Now we have to prove that δ > 0 x S x x 0 < δ and signx signx 0 ε We pick an arbitrary δ > 0 and we must show that x S x x 0 < δ and signx signx 0 ε In other words, we have to find some x S with the property x x 0 < δ and signx signx 0 ε. We evaluate the data that we know: x 0 = 0, sign0 = 0 and ε = 1. Hence we have to find some x S with the property x < δ and signx 1. But this is easy: Any x R with x 0, of absolute value < δ satisfies this property. f It seems plausible to claim that most functions are continuous. However, strictly speaking the opposite is true; it s even worse, there are many functions R R which are not continuous at any point x 0 R. Here is an example. Look at the so-called characteristic function f : R R of Q, defined by { 1 if x Q fx = 0 if x Q. f is not continuous at any x 0 R see example sheet 6.